Applications: elevation/depression, bearings practice questions.

From a point on level ground 40 m from the base of a vertical tower, the angle of elevation of the top of the tower is 45°. Determine the height of the tower.

• 20 m
• 40√2 m
• 40 m
• 40√3 m

Answer: 40 m

tan 45° = height ÷ 40 and tan 45° = 1, so the height equals the horizontal distance: 40 m.

A ship sails 6 km on a bearing of 090°, then 8 km on a bearing of 180°. Calculate the direct distance of the ship from its starting point.

• 10 km
• 14 km
• 2 km
• 2√7 km

Answer: 10 km

The legs are due east and due south, so they are perpendicular; by Pythagoras √(6² + 8²) = √100 = 10 km.

From the top of a vertical cliff 30 m high, the angle of depression of a boat at sea is 30°. Determine the horizontal distance from the boat to the base of the cliff.

• 30 m
• 10√3 m
• 15√3 m
• 30√3 m

Answer: 30√3 m

The angle of elevation from the boat is also 30°, so tan 30° = 30 ÷ d, giving d = 30 ÷ (1/√3) = 30√3 m.

The bearing of town B from town A is 060°. State the bearing of town A from town B.

• 120°
• 240°
• 300°
• 060°

Answer: 240°

For a back-bearing, add 180° when the original bearing is less than 180°: 060° + 180° = 240°.

A surveyor stands 20 m from the base of a flagpole and measures the angle of elevation of its top as 60°. Calculate the height of the flagpole.

• 20√3 m
• 20/√3 m
• 10√3 m
• 40 m

Answer: 20√3 m

Height = 20 × tan 60° = 20√3 m; a common error is dividing by tan 60° instead of multiplying.

A yacht leaves harbour and sails 10 km on a bearing of 030°. Determine how far north of the harbour the yacht is.

• 5 km
• 10√3 km
• 5√3 km
• 10 km

Answer: 5√3 km

The northward component is 10 cos 30° = 10 × (√3/2) = 5√3 km; the eastward component would be 10 sin 30° = 5 km.